Question

The equivalent mass of $H_{3}P{O}_4$ in the reaction, $H_{3}P{O}_4+NaOH\to Na{H}_{2}P{O}_4+H_{2}O$, is:

• A. 49
• B. 98
• C. 32.6
• D. 40

Answer 1

Answer: (B)

98

$H_{3}P{O}_4+NaOH⟶Na{H}_{2}P{O}_4+H_{2}O$

Number of $H^{+}$ ions released by $H_{3}P{O}_4$ in the reaction is $1$.

$\therefore$ Equivalent mass of $H_{3}P{O}_4$ is $\frac{\text{Molecular mass}}{1}=\frac{98}{1}=98$.